The inverse function theorem

Try

The map F(x,y)=(x+y,x−y) has invertible derivative everywhere and an explicit linear inverse. A nonlinear map with invertible derivative at one point resembles such a map sufficiently closely nearby. The inverse function theorem turns that resemblance into a local C¹ inverse. It does not say the map is globally injective or that a numerical Newton iteration works from every starting point.

Predict

Predict why the derivative must be invertible in equal dimensions. A singular linear approximation loses at least one direction, so it cannot be inverted as a linear map. The nonlinear function may still have a continuous inverse, as x³ does, but its inverse need not be differentiable. The theorem’s hypothesis is sufficient for a strong smooth inverse conclusion, not necessary for every weaker notion of invertibility.

Experiment

Normalize the local problem.

Step	Result
Translate a to zero	Base input is 0
Translate F(a) to zero	Base output is 0
Multiply output by DF(a)⁻¹	Derivative at zero becomes I
Seek F(x)=y	Fixed point of T_y(x)=y+x−F(x)

These invertible affine changes preserve the local question. In the normalized coordinates, the difference between F and the identity has a small derivative near zero, creating the contraction.

Understand

Choose a closed radius-r ball inside the domain on which ||I−DF||≤q<1; derivative continuity permits this. The segment estimate gives ||(x−F(x))−(z−F(z))||≤q||x−z||. Thus T_y is a contraction. If ||y||<(1−q)r, then ||T_y(x)||≤||y||+q||x||≤r, so it maps the complete closed ball into itself. Banach’s theorem supplies a unique solution x to F(x)=y, and the strict bound makes ||x||<r.

Also ||F(x)−F(z)||≥(1−q)||x−z|| on the ball, proving injectivity there and a Lipschitz bound for the inverse. Choose a small open output ball W and let V be its preimage inside the open input ball. Then V is open, contains zero, and F:V→W is bijective. This explicitly provides the neighborhoods in the theorem, rather than confusing the closed contraction ball with the final open domain.

Apply

To prove inverse differentiability at y=F(x), write an output increment k=F(x+h)−F(x)=Ah+r(h), with A=DF(x) invertible and r=o(||h||). The inverse Lipschitz estimate gives ||h||≤C||k||. Rearranging yields h=A⁻¹k−A⁻¹r(h), whose remainder is o(||k||). Thus the inverse derivative is A⁻¹. Its continuity follows from continuity of DF, continuity of the inverse map and continuity of matrix inversion on invertible matrices.

Repair a global conclusion using F(x,y)=(e^x cos y,e^x sin y). Its determinant is e^(2x)>0 everywhere, but adding 2π to y leaves the output unchanged. Each suitable neighborhood has an inverse branch; the whole plane does not admit one global inverse. Near the normalized base point, invertibility of each derivative follows from ||I−DF||<1: a zero image would imply ||v||≤q||v||, forcing v=0; a square injective matrix is invertible. Matrix inversion is continuous there through the adjugate-over-determinant formula.

Continue

Optional depth connects the fixed-point construction with Newton-like methods. The proof uses a fixed normalization matrix near the base point, not an unqualified guarantee for arbitrary Newton steps. Invariant neighborhoods and contraction constants remain essential.

State the theorem precisely: a C¹ map on an open subset of R^d with invertible derivative at a restricts to a C¹ bijection between neighborhoods of a and F(a), with the derivative formula above. Save the self-map estimate, injectivity estimate and inverse remainder argument as separate parts. Next, a block-coordinate construction will turn this local inversion result into an implicit function theorem for constraints.